Introductory Multivariable Functions

**Jacobpm64** · August 23rd 2007, 03:47 PM

I just started a multivariable calculus class, and I have a few problems I could use some help with.

I will provide my work so that you people know what I'm thinking.

1.) The monthly payments, P dollars, on a mortgage in which A dollars were borrowed at an annual interest rate of r% for t years is given by $P = f(A,r,t)$ . Is f an increasing or decreasing function of A? Of r? Of t?

Is this question asking if A, r, and t are plotted separately against P, would the graph be increasing or decreasing?

If so, I got:
for A : Increasing
for r : Increasing
for t : Decreasing

2.)Describe the set of points whose distance from the x-axis equals the distance from the yz-plane.

I don't know exactly how to word this one. I think if I would draw the xyz coordinate axes, and label the origin O, then the answer would be the line exactly bisecting the angle XOY. Is this correct? (if anyone can understand what i'm saying)

3.)Find the equations of planes that just touch the sphere $(x-2)^2 + (y-3)^2 + (z-3)^2 = 16$ and are parallel to:
(a) the xy-plane
(b) the yz-plane
(c) the xz-plane

(a) Since it has to be parallel to the xy-plane, we just deal with the z-coordinate.. so we have to subtract/add the radius to the z-coordinate of the center of the sphere in order to get the plane to just touch the sphere. So I think this would be z = 7 and z =-1.
(b) x = 6 and x = -2 (by the same reasoning)
(c) y = 7 and y = -1 (by the same reasoning)

I'm not sure if all of this is correct or not (it might be). I just need some reassurance since this is my first time working in 3-dimensions, and I find the transition a little confusing.

**ThePerfectHacker** · August 23rd 2007, 03:57 PM

Originally Posted by Jacobpm64

2.)Describe the set of points whose distance from the x-axis equals the distance from the yz-plane.

Let $(x,y,z)$ be a point.

1)Compute the distance from $(x,y,z)$ to xy-plane (which is) $1x+0y+0z = 0$ .

2)Compute the distance from $(x,y,z)$ to the x-axis (which is) $x=t,y=0,z=0$ .

Now equate #2 and #3.

**ThePerfectHacker** · August 23rd 2007, 04:07 PM

Originally Posted by Jacobpm64

3.)Find the equations of planes that just touch the sphere $(x-2)^2 + (y-3)^2 + (z-3)^2 = 16$ and are parallel to:
(a) the xy-plane
(b) the yz-plane
(c) the xz-plane

The tangent plane to a sphere is perpendicular to the radius.

Let $(x_0,y_0,z_0)$ be a point on the sphere such that condition (a) is satisfied.

That means, $(x_0-2)^2+(y_0-3)^2+(z_0-3)^2 = 16$ , since it is on the sphere.

It also means the radius vector is $(x_0-2)\bold{i}+(y_0-3)\bold{j}+(z_0-3)\bold{k}$ . And this is perpendicular to the tangent plane. So it means this vector is the normal of the tangent plane.

Now if we want this to be parallel with (a), the xy-plane, then we need the tangent vector to be parallel with the tangent vector of the xy-plane (which is $\bold{k}$ ). That means $(x_0-2)\bold{i}+(y_0-3)\bold{j}+(z_0-3)\bold{k} = k(\bold{k})$
So, $x_0 = 2, y_0 = 3, z_0 = k+3$ .

If you subsitute that into the sphere equation we get,
$(k+3-3)^2 = 16 \implies k = \pm 4$ .

So $(2,3,-1)\mbox{ or }(2,3,7)$ . At these point the tangent plane is parallel with the xy-plane.

Try the next ones.

**Jacobpm64** · August 23rd 2007, 04:21 PM

Originally Posted by ThePerfectHacker

The tangent plane to a sphere is perpendicular to the radius.

Let $(x_0,y_0,z_0)$ be a point on the sphere such that condition (a) is satisfied.

That means, $(x_0-2)^2+(y_0-3)^2+(z_0-3)^2 = 16$ , since it is on the sphere.

It also means the radius vector is $(x_0-2)\bold{i}+(y_0-3)\bold{j}+(z_0-3)\bold{k}$ . And this is perpendicular to the tangent plane. So it means this vector is the normal of the tangent plane.

Now if we want this to be parallel with (a), the xy-plane, then we need the tangent vector to be parallel with the tangent vector of the xy-plane (which is $\bold{k}$ ). That means $(x_0-2)\bold{i}+(y_0-3)\bold{j}+(z_0-3)\bold{k} = k(\bold{k})$
So, $x_0 = 2, y_0 = 3, z_0 = k+3$ .

If you subsitute that into the sphere equation we get,
$(k+3-3)^2 = 16 \implies k = \pm 4$ .

So $(2,3,-1)\mbox{ or }(2,3,7)$ . At these point the tangent plane is parallel with the xy-plane.

Try the next ones.

Okay, I understand, but my question asks for the equations of planes that are tangent to the sphere (not the points where they are tangent).

are my answers correctly written to be equations of planes?

(the vectors and stuff aren't expected of me yet, as we are only on the first section of the book.. vectors is later in the book).

**Jacobpm64** · August 23rd 2007, 04:53 PM

I'm still having trouble with #2.

I can't visualize it for anything.

The best I can describe it.. is if i was looking at the corner of a room, the answer would be a line coming out of the bottom corner running straight along the floor in such a way to bisect the angle made by the intersection of the walls.

**ThePerfectHacker** · August 23rd 2007, 06:27 PM

Originally Posted by Jacobpm64

2.)Describe the set of points whose distance from the x-axis equals the distance from the yz-plane.

Let $(x_0,y_0,z_0)$ be a point in space.

Here are two formulas we need to know:
1)Given a plane $Ax+By+Cz=D$ and a point $(x_0,y_0,z_0)$ the distance is given by, $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2+C^2}}$

2)Given a line $\ell$ and a point $P$ . The distance between between them is $\frac{||\bold{r}\times \bold{F}||}{||\bold{F}||}$ . Where $\bold{r}$ is any vector drawn from $P$ to $\ell$ . And $\bold{F}$ is any non-zero vector aligned with $\ell$ .

Distance to Plane: The equation of the yz-plane is given by $1x+0y+0z=0$ . So the distance is given by $\frac{|x_0|}{\sqrt{1}} = |x_0|$ (which is easy to see geometrically).

Distance to Line: So we need to find any $\bold{F} \mbox{ and }\bold{r}$ . We can pick $\bold{F} = \bold{i}$ . And let $\bold{r}$ be the vector from $(x_0,y_0,z_0)$ to the origin (which is on the line). Thus, $\bold{r} = x_0\bold{i}+y_)\bold{j}+z_0\bold{k}$ .
Compute the cross product, $\bold{r}\times \bold{F} = z_0\bold{j}-y_0\bold{k}$ . Thus, $|| \bold{r}\times \bold{F} || = \sqrt{y_0^2+z_0^2}$ . So the distance is given by, $\frac{\sqrt{y_0^2+z_0^2}}{1} = \sqrt{y_0^2+z_0^2}$ .

Which means that, $|x_0| = \sqrt{y_0^2+z_0^2}$ , square both sides (you are not losing anything because of absolute value) and we get $x_0^2 = y_0^2+z_0^2$ .

Now, if you know your quadric surfaces you should immediately realize this is a cone opening along the x-axis.

**Jacobpm64** · August 23rd 2007, 06:29 PM

Thanks a lot.

I got it now.

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